Solve Using the Quadratic Formula 6x^2-3x=0

$6 x^{2} - 3 x = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 6$ , $b = - 3$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (6 \cdot 0)}}{2 \cdot 6}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 6 \cdot 0}}{2 \cdot 6}$

Multiply $- 4 \cdot 6 \cdot 0$ .

Multiply $- 4$ by $6$ .

$x = \frac{3 \pm \sqrt{9 - 24 \cdot 0}}{2 \cdot 6}$

Multiply $- 24$ by $0$ .

$x = \frac{3 \pm \sqrt{9 + 0}}{2 \cdot 6}$

Add $9$ and $0$ .

$x = \frac{3 \pm \sqrt{9}}{2 \cdot 6}$

Rewrite $9$ as $3^{2}$ .

$x = \frac{3 \pm \sqrt{3^{2}}}{2 \cdot 6}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{3 \pm 3}{2 \cdot 6}$

Multiply $2$ by $6$ .

$x = \frac{3 \pm 3}{12}$

Simplify $\frac{3 \pm 3}{12}$ .

$x = \frac{1 \pm 1}{4}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{1}{2}, 0$